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Based on the data given below, the corre...

Based on the data given below, the correct order of reducing power is:
`Fe_((aq.))^(3+) + e rarr Fe_((aq.))^(2+), E^(@) = +0.77 V`
`Al_((aq.))^(3+) + 3e rarr Al_((s)), E^(@) = -1.66 V`
`Br_(2(aq.)) + 2e rarr 2Br_((aq.))^(-), E^(@) = +1.08 V`

A

`Fe^(2+)` is stronger reducing agent than `Br^(-)`

B

`Fe^(2+)` is stronger oxidising agent than Al

C

All is stronger reducing agent than `Fe^(2+)`

D

`Br^(-)` is stronger reducing agent than Al

Text Solution

AI Generated Solution

The correct Answer is:
To determine the correct order of reducing power based on the given standard reduction potentials (E° values), we will analyze the data step by step. ### Step 1: Understand the Concept of Reducing Power Reducing power is the ability of a species to donate electrons and reduce another species. The standard reduction potential (E°) indicates the tendency of a species to gain electrons. A lower (more negative) E° value indicates a stronger reducing agent. ### Step 2: Analyze the Given Reactions and E° Values We have three half-reactions with their respective E° values: 1. \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}, \quad E° = +0.77 \, \text{V} \) 2. \( \text{Al}^{3+} + 3e^- \rightarrow \text{Al}, \quad E° = -1.66 \, \text{V} \) 3. \( \text{Br}_2 + 2e^- \rightarrow 2\text{Br}^-, \quad E° = +1.08 \, \text{V} \) ### Step 3: Determine the Reducing Power - For \( \text{Fe}^{3+}/\text{Fe}^{2+} \): E° = +0.77 V (less reducing power) - For \( \text{Al}^{3+}/\text{Al} \): E° = -1.66 V (more reducing power) - For \( \text{Br}_2/\text{Br}^- \): E° = +1.08 V (less reducing power) ### Step 4: Order the Reducing Agents - **Aluminum (Al)** has the lowest E° value (-1.66 V), making it the strongest reducing agent. - **Iron (Fe)** comes next with an E° of +0.77 V. - **Bromine (Br)** has the highest E° value (+1.08 V), making it the weakest reducing agent. ### Conclusion: Correct Order of Reducing Power The order of reducing power from strongest to weakest is: 1. \( \text{Al} \) 2. \( \text{Fe}^{2+} \) 3. \( \text{Br}^- \) ### Final Answer The correct order of reducing power is: \[ \text{Al} > \text{Fe}^{2+} > \text{Br}^- \] ---
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Standard electrode potential data is given below : Fe^(3+) (aq) + e ^(-) rarr Fe^(2+) (aq) , E^(o) = + 0.77 V Al^(3+) (aq) + 3e^(-) rarr Al (s) , E^(o) = - 1.66 V Br_(2) (aq) + 2e^(-) rarr 2Br^(-) (aq) , E^(o) = +1.08 V Based on the data given above, reducing power of Fe^(2+) Al and Br^(-) will increase in the order :

Based on data given below for the electrode potential , reducing power of Fe^(2+) , Al and Br^(-) will increase in the order Br_(2)(aq)+2e^(-)rarr2Br^(-)(aq),E^@=+1.08V Al^(3+)(aq)+3e^(-)rarrAl(s),E^@=-1.66V Fe^(3+)(aq)+e^(-2)rarrFe^(+2) (aq) , E^(@) =+0.77V

Knowledge Check

  • Based on the data given below, the correcy order of reducing power is: Fe_((aq.))^(3+) + e rarr Fe_((aq.))^(2+), E^(@) = +0.77 V Al_((aq.))^(3+) + 3e rarr Al_((s)), E^(@) = -1.66 V Br_(2(aq.)) + 2e rarr 2Br_((aq.))^(-), E^(@) = +1.08 V

    A
    `Br^(-) lt Fe^(2+) lt Al`
    B
    `Fe^(2+) lt Al lt Br^-`
    C
    `Al lt Br^(-) lt Fe^(2+)`
    D
    `Al lt Fe^(2+) lt Br^(-)`

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